Evaluating Trigonometric Expressions with cot A

The problem asks us to find the value of the expression \((\sin A + \cos A) \times \text{cosec } A\) given that \(\cot A = \frac{12}{5}\). We can solve this problem using trigonometric identities, which is often the most efficient way.

Using Trigonometric Identities to Simplify

We are given the expression \((\sin A + \cos A) \times \text{cosec } A\).

We know that \(\text{cosec } A\) is the reciprocal of \(\sin A\), i.e., \(\text{cosec } A = \frac{1}{\sin A}\).

Let's distribute \(\text{cosec } A\) across the terms inside the parenthesis:

• \((\sin A + \cos A) \times \text{cosec } A = \sin A \times \text{cosec } A + \cos A \times \text{cosec } A\)

Substitute \(\text{cosec } A = \frac{1}{\sin A}\) into the expression:

• \(= \sin A \times \frac{1}{\sin A} + \cos A \times \frac{1}{\sin A}\)

The first term simplifies:

• \(= 1 + \frac{\cos A}{\sin A}\)

We also know that \(\cot A\) is defined as the ratio of \(\cos A\) to \(\sin A\), i.e., \(\cot A = \frac{\cos A}{\sin A}\).

So, the expression simplifies to:

• \(= 1 + \cot A\)

We are given that \(\cot A = \frac{12}{5}\). Substituting this value into the simplified expression:

• \(= 1 + \frac{12}{5}\)

To add these values, find a common denominator:

• \(= \frac{5}{5} + \frac{12}{5}\)
• \(= \frac{5+12}{5}\)
• \(= \frac{17}{5}\)

Thus, the value of \((\sin A + \cos A) \times \text{cosec } A\) is \(\frac{17}{5}\).

Alternative Method: Using a Right-Angled Triangle

We can also solve this by constructing a right-angled triangle. Given \(\cot A = \frac{12}{5}\). In a right-angled triangle, \(\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}}\).

Assume the adjacent side is \(12k\) and the opposite side is \(5k\) for some positive constant \(k\).

Using the Pythagorean theorem, the hypotenuse (\(h\)) is:

\(h^2 = (12k)^2 + (5k)^2\) \(h^2 = 144k^2 + 25k^2\) \(h^2 = 169k^2\) \(h = \sqrt{169k^2} = 13k\)

Now we can find \(\sin A\), \(\cos A\), and \(\text{cosec } A\) using the triangle's sides:

• \(\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{5k}{13k} = \frac{5}{13}\)
• \(\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{12k}{13k} = \frac{12}{13}\)
• \(\text{cosec } A = \frac{1}{\sin A} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\)

Now substitute these values into the expression \((\sin A + \cos A) \times \text{cosec } A\):

• \((\frac{5}{13} + \frac{12}{13}) \times \frac{13}{5}\)
• \((\frac{5+12}{13}) \times \frac{13}{5}\)
• \((\frac{17}{13}) \times \frac{13}{5}\)
• \(= \frac{17}{5}\)

Both methods yield the same result.

Summary of Calculations

Here is a summary of the steps using the identity method:

The final value of the expression \((\sin A + \cos A) \times \text{cosec } A\) is \(\frac{17}{5}\).

Revision Table: Key Trigonometric Ratios & Identities

Additional Information: Understanding Reciprocal Identities

The problem heavily relied on the reciprocal identities in trigonometry. These identities relate pairs of trigonometric functions that are reciprocals of each other.

• \(\sin \theta\) and \(\text{cosec } \theta\): \(\text{cosec } \theta = \frac{1}{\sin \theta}\) or \(\sin \theta \times \text{cosec } \theta = 1\).
• \(\cos \theta\) and \(\text{sec } \theta\): \(\text{sec } \theta = \frac{1}{\cos \theta}\) or \(\cos \theta \times \text{sec } \theta = 1\).
• \(\tan \theta\) and \(\cot \theta\): \(\cot \theta = \frac{1}{\tan \theta}\) or \(\tan \theta \times \cot \theta = 1\).

Understanding these fundamental identities allows for simplification of complex trigonometric expressions, making calculations much easier, as demonstrated in this problem. The quotient identity \(\cot A = \frac{\cos A}{\sin A}\) was also crucial for the identity-based solution.